# Functions with formula like an exponential function

I read the answers to What are argument one can give to students on the definition $0^0$? and I noticed some people are mildly bothered that $0^x$ is discontinuous if we define $0^0=1$. However, I think is fairly obvious that if $f(x)=0^x$ then $\text{domain}(f)$ does not contain negative numbers. Surely, if $x <0$ then $$0^x = 0^{-(-x)} = \frac{1}{0^{-x}}=\frac{1}{0}$$ usually an unwelcome expression. My thought here, $0^x$ is not a proper exponential function because the natural domain of the expression $0^x$ is at largest $[0, \infty)$ and, given the convention $0^0=1$, is continuous on $(0,\infty)$. In other words, it is the constant function $g(x)=0$ restricted to $(0,\infty)$. On the other hand, exponential functions with $a>0$ and $a \neq 1$ all have nice properties: they're injective functions with $\text{range}(a^x) = (0, \infty)$ and $\text{domain}(a^x) = (-\infty, \infty)$. Of course, $a=1$ is somewhat boring $a^x=1^x=1$ hence we just have the constant function with an awkward formula.

Ok, enough about the happy place. What then about $f(x)=a^x$ where $a<0$. What is the natural domain for this formula? This leads to my question:

Do you mention the problem of understanding $f(x)=a^x$ for $a<0$ in your teaching?
Is it wise to omit discussion of $a \leq 0$ ?

I don't have a firm opinion on this, I usually find myself talking about the bizarre behavior of such a formula sometime in the first week of calculus. A secondary question, does appreciating the weirdness of negative base "exponential functions" help explain why we shouldn't have much hope for $0^x$ to behave nicely? I put quotes here because I think the definition of an exponential function properly assumes the base is a positive number which is not one.

• In Germany, this is part of secondary education, not undergraduate education. – Toscho Jun 20 '14 at 10:19
• A proper answer would depend on how you define the function $b^x$ where $b \in \mathbb{R}_+$ and $x\in \mathbb{R}$. (a) you can define it by first defining it for $x\in \mathbb{N}$, followed by $x\in \mathbb{Q}$, and then extending to $\mathbb{R}$ by continuity. (b) You can also "define" $b^x := \exp( \ln(b) x)$. The case $b < 0$ in case (a) requires finding (among other things) $b^{-1/2n}$ which leads naturally to complex numbers. In case (b) you need to consider $\ln(b)$ for $b < 0$... which you can of course also deal with complex analysis if you allow multiple-valued functions. – Willie Wong Jun 20 '14 at 12:34
• My point is that some avenues of approach naturally leads to the discussion of branch cuts in complex analysis, which is a useful but deep subject. If you feel comfortable going the whole way: sure, go ahead. But if you don't feel comfortable showing them the whole story, maybe it is easier to just not confuse them needlessly. – Willie Wong Jun 20 '14 at 12:35
• @Toscho I added the secondary education tag in view of the international nature of the MESE :) – James S. Cook Jun 21 '14 at 1:46

When you cover exponential functions, you should always discuss negative bases and base 0. But using power laws, it's relatively easy to see, that $(-2)^x$ is problematic for non-integer $x$, and $0^x$ is problematic for negative $x$. If you use permanence sequences, you can also easily see, that $0^x$ is also problematic for $x=0$.